James K. Peterson Department of Biological Sciences and Department of Mathematical Sciences Clemson University August 30, 2017
Outline 1 2 s 3 Basic Results 4 Homework
Let S be a set of real numbers. We need to make precise the idea of a set of real numbers being bounded. Definition We say a set S is bounded above if there is a number M so that x M for all x in S. We call M an upper bound of S or just an u.b. If S = {y : y = x 2 and 1 x 2}, there are many u.b. s of S. Some choices are M = 5, M = 4.1. Note M = 1.9 is not an u.b. If S = {y : y = tanh(x) and x R}, there are many u.b. s of S. Some choices are M = 2, M = 2.1. Note M = 0 is not an u.b. Draw a picture of this graph too.
Let S be a set of real numbers. Definition We say a set S is bounded below if there is a number m so that x m for all x in S. We call m a lower bound of S or just a l.b. If S = {y : y = x 2 and 1 x 2}, there are many l.b. s of S. Some choices are m = 2, m = 0.1. Note m = 0.3 is not a l.b. If S = {y : y = tanh(x) and x R}, there are many l.b. s of S. Some choices are m = 1.1, m = 1.05. Note m = 0.87 is not a l.b. Draw a picture of this graph again.
Let S be a set of real numbers. Definition We say a set S is bounded if S is bounded above and bounded below. That is, there are finite numbers m and M so that m x M for all x S. We usually overestimate the bound even more and say B is bounded if we can find a number B so that x B for all x S. A good choice of such a B is to let B = max( m, M ) for any choice of l.b. m and u.b. M. If S = {y : y = x 2 and 1 x < 2}. here S = [0, 4) and so for m = 2 and M = 5, a choice of B is B = 5. Of course, there are many other choices of B. If S = {y : y = tanh(x) and x R}, we have S = ( 1, 1) and for m = 1.1 and M = 1.2, a choice of B is B = 1.2.
The next material is more abstract! We need to introduce the notion of least upper bound and greatest lower bound. We also call the least upper bound the l.u.b.. It is also called the supremum of the set S. We use the notation sup(s) as well. We also call the greatest lower bound the g.l.b.. It is also called the infimum of the set S. We use the notation inf(s) as well. Definition The least upper bound, l.u.b. or sup of the set S is a number U satisfying 1 U is an upper bound of S 2 If M is any other upper bound of S, then U M. The greatest lower bound, g.l.b. or inf of the set S is a number u satisfying 1 u is a lower bound of S 2 If m is any other lower bound of S, then u m.
If S = {y : y = x 2 and 1 x < 2}. here S = [0, 4) and so inf(s) = 0 and sup(s) = 4. If S = {y : y = tanh(x) and x R}, we have inf (S) = 1 and sup(s) = 1. Not the inf and sup of a set S need NOT be in S! If S = {y : cos(2nπ/3), n N}, The only possible values in S are cos(2π/3) = 1/2, cos(4π/3) = 1/2 and cos(6π/3) = 1. There are no other values and these 2 values are endlessly repeated in a cycle. Here inf(s) = 1/2 and sup(s) = 1.
Comment If a set S has no finite lower bound, we set inf(s) =. If a set S has no finite upper bound, we set sup(s) =. Comment If the set S =, we set inf(s) = and sup(s) =. Definition We say Q S is a maximum of S if sup(s) = Q. This is the same, of course, as saying x Q for all x in S which is the usual definition of an upper bound. But this is different as Q is in S. We call Q a maximizer or maximum element of S. We say q S is a minimum of S if inf(s) = q. Again, this is the same as saying x q for all x in S which is the usual definition of a lower bound. But this is different as q is in S. We call q a minimizer minimal element of S.
There is a fundamental axiom about the behavior of the real numbers which is very important. Axiom The Completeness Axiom Let S be a set of real numbers which is nonempty and bounded above. Then the supreumum of S exists and is finite. Let S be a set of real numbers which is nonempty and bounded below. Then the infimum of S exists and is finite. Comment So nonempty bounded sets of real numbers always have a finite infimum and supremum. This does not say the set has a finite minimum and finite maximum. Another way of saying this is that we don t know if S has a minimizer and maximizer.
Theorem Let S be a nonempty set of real numbers which is bounded above. Then sup(s) exists and is finite. Then S has a maximal element if and only if (IFF) sup(s) S. Proof ( ): Assume sup(s) is in S. By definition, sup(s) is an upper bound of S and so must satisfy x sup(s) for all x in S. This says sup(s) is a maximizer of S. ( ): Let Q denote a maximizer of S. Then by definition x Q for all x in S and is an upper bound. So by the definition of a supremum, sup(s) Q. Since Q is a maximizer, Q is in S and from the definition of upper bound, we have Q sup(s) as well. This says sup(s) Q sup(s) or sup(s) = Q.
Theorem Let S be a nonempty set of real numbers which is bounded below. Then inf(s) exists and is finite. Then S has a minimal element inf(s) S. Proof ( ): Assume inf(s) is in S. By definition, inf(s) is a lower bound of S and so must satisfy x inf(s) for all x in S. This says inf(s) is a minimizer of S. ( ): Let q denote a minimizer of S. Then by definition x q for all x in S and is a lower bound. So by the definition of an infimum, q inf(s). Since q is a minimizer, q is in S and from the definition of lower bound, we have inf(s) q as well. This says inf(s) q inf(s) or inf(s) = q.
Lemma Infimum Tolerance Lemma: Let S be a nonempty set of real numbers that is bounded below. Let ɛ > 0 be arbitrarily chosen. Then y S inf(s) y < inf(s) + ɛ Proof We do this by contradiction. Assume this is not true for some ɛ > 0. Then for all y in S, we must have y inf(s) + ɛ. But this says inf(s) + ɛ must be a lower bound of S. So by the definition of infimum, we must have inf(s) inf(s) + ɛ for a positive epsilon which is impossible. Thus our assumption is wrong and we must be able to find at least one y in S that satisfies inf(s) y < inf(s) + ɛ.
Lemma Supremum Tolerance Lemma: Let S be a nonempty set of real numbers that is bounded above. Let ɛ > 0 be arbitrarily chosen. Then y S sup(s) ɛ < y sup(s) Proof We do this by contradiction. Assume this is not true for some ɛ > 0. Then for all y in S, we must have y sup(s) ɛ. But this says sup(s) ɛ must be an upper bound of S. So by the definition of supremum, we must have sup(s) sup(s) ɛ for a positive epsilon which is impossible. Thus our assumption is wrong and we must be able to find at least one y in S that satisfies sup(s) ɛ < y sup(s).
s Let f (x, y) = x + 2y and let S = [0, 1] [1, 3] which is also S x S y where S x = {x : 0 x 1} and S y = {y : 1 y 3}. Note inf (x,y) [0,1] [1,3] f (x, y) = 0 + 2 = 2 and sup (x,y) [0,1] [1,3] f (x, y) = 1 + 6 = 7. sup 0 x 1 inf 1 y 3 so in this example, inf f (x, y) 1 y 3 = inf (x + 2y) = x + 2 1 y 3 sup 0 x 1 f (x, y) = sup 0 x 1 (x + 2y) = 1 + 2y inf (x + 2y) = sup (x + 2) = 3 1 y 3 0 x 1 sup (x + 2y) = inf (1 + 2y) = 3 0 x 1 1 y 3. inf y S y sup x S x f (x, y) = sup x S x inf f (x, y) y S y
s Please look the text where you can see this function drawn out. That might help. Let 0, (x, y) (1/2, 1] (1/2, 1] f (x, y) = 2, (x, y) (1/2, 1] [0, 1/2] and [0, 1/2] (1/2, 1] 1, (x, y) [0, 1/2] [0, 1/2] and let S = [0, 1] [0, 1] which is also S x S y where S x = {x : 0 x 1} and S y = {y : 0 y 1}. Note inf (x,y) [0,1] [0,1] f (x, y) = 0 and sup (x,y) [0,1] [0,1] f (x, y) = 2. Then, we also can find inf f (x, y) = 0 y 1 { 1, 0 x 1/2 0, 1/2 < x 1 and sup f (x, y) = 0 x 1 { 2, 0 y 1/2 2, 1/2 < y 1
s (Continued) and sup 0 x 1 inf 0 y 1 so in this example { 1, 0 x 1/2 inf f (x, y) = sup 0 y 1 0 x 1 0, 1/2 < x 1 = 1. { 2, 0 y 1/2 sup f (x, y) = inf 0 x 1 0 y 1 2, 1/2 < y 1 = 2 inf y S y sup x S x f (x, y) sup x S x inf f (x, y) y S y and in fact sup x S x inf f (x, y) < inf y S y y S y sup f (x, y) x S x
s The moral here is that order matters. For example, in an applied optimization problem, it is not always true that min x max y f (x, y) = max y where x and y come from some domain set S. min x f (x, y)
s The moral here is that order matters. For example, in an applied optimization problem, it is not always true that min x max y f (x, y) = max y where x and y come from some domain set S. min x f (x, y) So it is probably important to find out when the order does not matter because it might be easier to compute in one ordering than another.
Basic Results Theorem Let S be a nonempty bounded set of real numbers. Then inf(s) and sup(s) are unique. Proof By the completeness axiom, since S is bounded and nonempty, we know inf(s) and sup(s) are finite numbers. Let u 2 satisfy the definition of supremum also. Then, we know u 2 M for all upper bounds M of S and in particular since sup(s) is an upper bound too, we must have u 2 sup(s). But since sup(s) is a supremum, by definition, we also know sup(s) u 2 as u 2 is an upper bound. Combining, we have u 2 sup(s) u 2 which tells us u 2 = sup(s). A similar argument shows the inf(s) is also unique.
Homework Homework 4 4.1 Let f (x, y) = 3, (x, y) (1/2, 1] (1/2, 1] 2, (x, y) (1/2, 1] [0, 1/2] and [0, 1/2] (1/2, 1] 4, (x, y) [0, 1/2] [0, 1/2] and let S = [0, 1] [0, 1] which is also S x S y where S x = {x : 0 x 1} and S y = {y : 0 y 1}. Find 1 inf (x,y) S f (x, y), and sup (x,y) S f (x, y), 2 inf y Sy sup x Sx f (x, y) and sup x Sx inf y Sy f (x, y). 3 inf x Sx sup y Sy f (x, y) and sup y Sy inf x Sx f (x, y). 4.2 Let S = {z : z = e x 2 y 2 for (x, y) R 2 }. Find inf(s) and sup(s). Does the minimum and maximum of S exist and if so what are their values?